Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $a \neq 0$. $z = \dfrac{-10}{16a + 28} \times \dfrac{10(4a + 7)}{2a} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $z = \dfrac{ -10 \times 10(4a + 7) } { (16a + 28) \times 2a } $ $ z = \dfrac {-10 \times 10(4a + 7)} {2a \times 4(4a + 7)} $ $ z = \dfrac{-100(4a + 7)}{8a(4a + 7)} $ We can cancel the $4a + 7$ so long as $4a + 7 \neq 0$ Therefore $a \neq -\dfrac{7}{4}$ $z = \dfrac{-100 \cancel{(4a + 7})}{8a \cancel{(4a + 7)}} = -\dfrac{100}{8a} = -\dfrac{25}{2a} $